Problem: Ikta has an extraordinary feeling for undirected graphs. Ikta likes the undirected graph G and its two-point s, t pair (G, s, t) that has the greatest "beauty". The "beauty" of a pair (G, s, t) is the edge e = \\ {u, v \\} (u and v are two different points of G), and the shortest path from s to t in G. Is the number of things whose length is greater than the length of the shortest path from s to t in the undirected graph with g plus e.

Your job is to write a program that seeks its "beauty" given a pair (G, s, t).



Input

The input is given in the following format.

> N M s t
> x1 y1
> ...
> xi yi
> ...
> xM yM
>

First, the number of vertices, the number of edges, and the integers N, M, s, t representing the two vertices of the undirected graph are input. From the 2nd line to the M + 1th line, two vertices connected by edges are input. (However, let the set of G vertices be \\ {1, ..., N \\}.)

Constraints

Each variable being input satisfies the following constraints.

* 2 ≤ N ≤ 100,000
* 1 ≤ M ≤ 300,000
* 1 ≤ s, t, xi, yi ≤ N
* Different from s and t
* Guaranteed to reach t from s

Output

When the given graph is G, output the "beauty" of the set (G, s, t) in one line.

Examples

Input

3 2 1 3
1 2
2 3


Output

1


Input

9 8 7 8
2 6
4 9
8 6
9 2
3 8
1 8
8 5
7 9


Output

7


Input

4 3 1 4
1 2
3 4
4 1


Output

0


Input

9 7 8 9
9 6
6 5
3 6
3 7
2 5
8 5
1 4


Output

2

Solution (in CPP):

Solution:
#include <iostream>
#include <fstream>
#include <cstdio>
#include <cmath>
#include <vector>
#include <cstring>
#include <string>
#include <set>
#include <map>
#include <stack>
#include <queue>
#include <algorithm>
using namespace std;
 
#define REP(i,n) for(int i=0; i<n; ++i)
#define FOR(i,a,b) for(int i=a; i<=b; ++i)
#define FORR(i,a,b) for (int i=a; i>=b; --i)
#define pi M_PI
 
typedef long long ll;
typedef vector<int> VI;
typedef vector<ll> VL;
typedef vector<VI> VVI;
typedef pair<int,int> P;
typedef pair<ll,ll> PL;

const int N = 100010;
VI e[N];
ll d1[N], d2[N], p1[N], p2[N];

void bfs(int s, ll d[], ll p[]){
    REP(i,N) d[i] = -1;
    d[s] = 0;
    queue<int> que;
    que.push(s);
    while (!que.empty()){
        int now = que.front();
        que.pop();
        p[d[now]]++;
        REP(i,e[now].size()){
            int next = e[now][i];
            if (d[next] >= 0) continue;
            d[next] = d[now] + 1;
            que.push(next);
        }
    }
}

int main() {
    int n, m, s, t;
    cin >> n >> m >> s >> t;
    s--;
    t--;
    REP(i,m){
        int x, y;
        scanf("%d %d", &x, &y);
        x--;
        y--;
        e[x].push_back(y);
        e[y].push_back(x);
    }
    bfs(s,d1,p1);
    bfs(t,d2,p2);
    int x = d1[t];

    ll ans = 0;
    REP(i,x-1){
        ans += p1[i] * p2[x-2-i];
    }

    cout << ans << endl;

    return 0;
}